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Appendix: III.

These values of y and D have been co-ordinated in Fig. 882, and
a curve has been drawn whose vertical'ordinates vary from 637 to
+ 2395. The point where the curve crosses the axis of D shews
D = 6 when y  o. The true answer is

D = 6*005 ins-

P. 438. Distribution of Shear stress in Beams.
Just as the summation of vertical forces (loads and reactions)
in a beam will give a curve of total vertical shear at every
section (p. 853), so the summation of horizontal forces (tensile
and compressive stresses) will produce a curve of horizontal
shear stress intensity; and, as the tensile stresses oppose the
compressive stresses, the greatest shear is at the neutral axis.
The summation (see p. 851) of the stress areas of p. 431 must
then proceed from the upper or lower fibre to a maximum at
the centre; and two examples have been given in Fig. 883 to shew
how this should be done. Again, any elemental cube must be
subject to vertical shear on its vertical sides, and horizontal shear
on its horizontal sides, the one stress being caused by the presence
of the other; hence the distribution of vertical and horizontal
shear stresses over a section are identical Finally the scale of
diagrams c c, Fig. 882, can easily be found by equating the area
abed to the total vertical shear at the given section; and db>
the maximum unit stress, is thereby arrived at